It would be fun to know how to pull the whole energy supply of a single AA battery for a one time operation, like making a huge spark, firing a photo flash or a very bright light.

What kind of electronics would be needed?

upd: discharging the battery should power some useful device (or be useful without a device), not just discharging for the sake of it.

  • 2
    \$\begingroup\$ Instant as in "plug battery in, battery instantly dead" or "plug battery in, store battery energy, use energy instantly"? \$\endgroup\$ – W5VO May 10 '12 at 0:00
  • \$\begingroup\$ Any way to fully use all of a battery's energy will do. If we should store it somewhere first, so be it. \$\endgroup\$ – user1306322 May 10 '12 at 0:26
  • 9
    \$\begingroup\$ A warning to anyone who actually wants to try discharging the power of an AA battery in a single spark: they may seem small and harmless, but an alkaline AA battery contains about the equivalent of a .458 Magnum's muzzle energy. Such a spark lasting 100ms is essentially a 70kW device, running somewhere around 330V / 220A. Be extremely careful - it can kill you. \$\endgroup\$ – Polynomial May 10 '12 at 7:56
  • 6
    \$\begingroup\$ putting the battery on fire often results in an explosion where most of the battery's energy is released. But this is not utterly useful except for the visual effect, pollution and possible injuries. \$\endgroup\$ – Stefan Paul Noack May 10 '12 at 8:26
  • \$\begingroup\$ The question is as unsolvable with present technology as capturing the energy in lightning for useful purposes. the Laws of Physics dictate the transmission impedance requirements. The path and load must match the source. Since the air arcs with a fusion path of negative resistance in air, and all known storage solutions are both positive resistance and have excess latency (Time constant). It is impossible. Ditto for batteries due to latency time constant alone. Proof is in Heavside Equations based on Maxwell's Laws and almost as long as Fermat's Proof. \$\endgroup\$ – Sunnyskyguy EE75 May 10 '12 at 14:02

There is no practical way to dump "all of the energy" in a flash. But you can discharge it slowly into an expensive UltraCap with many Farads of capacity and a few mΩ ESR and then dump the ultraCap in a flash enough to weld copper.

This is how electronic flash cameras work today, except you get many flashes per battery charge using just normal low ESR caps not "UltraCaps".

Is this for Sir Guy Fox day or just for a lark to raise some eyebrows in airport security and end up in Gitmo or are you competing with the Swiss in generating a Higgs Boson particle?

  • 6
    \$\begingroup\$ For science. Science is fun. \$\endgroup\$ – user1306322 May 10 '12 at 1:18
  • 2
    \$\begingroup\$ Burns are not. Just because Mythbusters blows things up all the time doesn't mean it's science. Be careful. \$\endgroup\$ – AngryEE May 10 '12 at 11:52
  • 2
    \$\begingroup\$ I don't think that Guy Fawkes received a knighthood for his endeavours :) \$\endgroup\$ – MikeJ-UK May 10 '12 at 12:11
  • 4
    \$\begingroup\$ @MikeJ-UK "Treason doth never prosper: what’s the reason? Why, if it prosper, none dare call it treason." -- Sir John Harrington (1561–1612) \$\endgroup\$ – Dan Neely May 10 '12 at 12:35
  • \$\begingroup\$ I must have confused this modern day action hero with another "guy" , Sir Guy Fairfax of Steeton \$\endgroup\$ – Sunnyskyguy EE75 May 10 '12 at 12:40

If you force a very high current (much higher than the short-circuit current) into the battery's negative node, and out of the positive node, you'll be able to remove all the energy in it in a minimum time, but that time cannot be zero.

Also, to be able to inject that current, you'll need to treat the battery as a load, and connect it, upside down, to a high voltage source or (preferably) a high current source. The battery won't like it, though.

Schematic 1

The internal resistance of a battery does not limit the current. That would be true only if all the voltage that we had access to was the one from the battery itself. But that is not true. No one impedes us from connecting an external voltage. The question asks for the fastest way to remove all possible energy from a battery.

In the following example, the short-circuit current of the battery would be 1.5/0.2=7.5 A. By connecting it upside down to a voltage generator with (for instance) 100 V and 0.01 ohm, we can force a current of (100+1.5)/(0.2+0.01)=483 A (in that ideal model), which is much higher than the short circuit current.

Schematic 2

If you slowly transfer the energy in a battery to a supercap, and then quickly consume the energy in the supercap, that is not answering the question. The question asks how can we use all the energy in a battery (not in a cap) in an instant.

  • \$\begingroup\$ Actually using a supercap is an allowable solution -- the OP even said so (before you added the paragraph about the supercap). They said, "Any way to fully use all of a battery's energy will do. If we should store it somewhere first, so be it. " \$\endgroup\$ – tcrosley May 10 '12 at 0:55
  • \$\begingroup\$ I meant some practical use indeed. Though simply discharging a battery is also an interesting matter, I wonder how it would react. \$\endgroup\$ – user1306322 May 10 '12 at 1:20
  • \$\begingroup\$ Technically if you are charging the cap from the battery without a switching power supply you will only be using half of the battery's energy in an instant. \$\endgroup\$ – charliehorse55 May 10 '12 at 5:33
  • \$\begingroup\$ @tcrosley That comment that the OP wrote is not coherent with his question (because it implies slowly removing the energy from the battery). If he wanted that comment to prevail, he should change the question. \$\endgroup\$ – Telaclavo May 10 '12 at 9:42
  • 3
    \$\begingroup\$ When you start talking about 500 amp currents, you need to talk to an electrochemist about the effect of turning the electrolyte to steam. There will be an optimum point where the electrolyte steams just as you pull the last erg out. Above that the steam explosion will scatter the anode and cathode before being completely reacted. What superficially appears to be a EE problem, is actually a chemE electrodynamics and thermodynamics problem. \$\endgroup\$ – Vince Mulhollon May 10 '12 at 11:50

You added:

Any way to fully use all of a battery's energy will do.
If we should store it somewhere first, so be it

Which makes it much easier, subject to conversion efficiencies.
A boost converter will help in some cases by maqtching thebattery output to the storage system.

All of the following rely on transferring the battery's energy to a system which allows rapid "discharge". All are potentially useful.

Death is possible with all of these with due carelessness.

About 10 kJ Joule of energy is typically available.
Enough to damage or kill in all cases.

  • Charge a supercap. When as much battery energy is transferred to cap as system uses allows. Discharge cap. A VERY large cap needed. Not really practical

  • Use electrolysis of water to produce a Hydrogen Oxygen mix. Ignite mixture. (Don't try this at home etc ... !!!). Eminently practical and potentially extremely dangerous.

  • Use a motor to spin up a flywheel. Apply the flywheel to "something". Easily enough doable.

  • Heat a Phase change metallic material with PC temperature of > 100 C (various Bismuth alloys meet this spec) so it liquidifies. Add water. Very doable. Finely divided matrix helps energy transfer to water. Potentially lethal.

More possible ...


Batteries have internal resistance. Since their output voltage is (for most intents and purposes) fixed, you have a fixed maximum current output. Telaclavo's answer covers a situation where this isn't true, but I'm going to walk you through a simple short of the battery.

A typical internal resistance for an AA battery is about 0.2Ω[1], so if we ignore the resistance of any connecting wires it's not too difficult to calculate the (theoretical) maximum output current. Since we know R and V, we can calculate I as I=V/R, therefore I = 1.5/0.2 = 7.5A maximum current[2].

Your average alkaline battery will give you about 2000mAh of juice under 0.5A load. I can't find any detailed specs on how long they last when shorted, but it's not going to be less than 1/10th that amount. So, at an absolute minimum, it's going to take about 90 seconds to drain the battery.

[1] 0.2Ω is what Duracell and Panasonic state are their "standard" internal resistances for "normal loads".
[2] There are other resistances involved, and the internal resistance changes as you increase the load on the battery. I'm just trying to keep things simple here.

  • \$\begingroup\$ I'm trying to understand if there is a way to charge some device that would then release accumulated energy very quickly. So I guess such device would be then charged at a limited rate, but how would the battery give away it's energy over time (what could happen to it, assuming it's a disposable one use power supply), and how said device should be made? \$\endgroup\$ – user1306322 May 10 '12 at 0:22
  • \$\begingroup\$ The flash circuits in disposable cameras use a capacitor with very low ESR to store a charge, then discharge that energy into the coil of a transformer when the button is pressed, essentially shorting the capacitor. The secondary coil of the transformer is wired to the flash, which receives a very quick jolt of current. If you've got enough capacitors in parallel, or a very expensive supercap, you can practically drain the entire battery out and do the same thing. \$\endgroup\$ – Polynomial May 10 '12 at 7:41
  • \$\begingroup\$ As a safety note, you're actually dealing with a lot of energy here. 2000mAh is 7200 joules, which is roughly equal to the muzzle energy of a .458 Magnum. If you discharge that in the form of a spark (roughly 100ms) that's power to the order of 70kW. Sparks usually run through air at about 330V, so you're looking at 215A. That's more than enough to kill you. Be very very careful. \$\endgroup\$ – Polynomial May 10 '12 at 7:51

It would be fun to know how to pull the whole energy supply of a single AA battery for a one time operation

Pedantic: Most of the energy in your battery is tied up in the nuclear forces that bind together the subatomic particles that make up the atoms in your battery. To liberate the whole supply of energy, you will need probably need to resort to nuclear fusion or fission.

Fun fact: There is more energy in the uranium impurities in a kilogram of coal than can be liberated by just burning that kilogram of coal!

  • \$\begingroup\$ Around 1kg of Uranium on average in soil you could excavate from a 1/4 acre section. Much more in some areas. Much less in others. \$\endgroup\$ – Russell McMahon May 10 '12 at 4:56
  • 4
    \$\begingroup\$ "To liberate the whole supply of energy, you will need probably need to resort to ... " antimatter. You're not going to get much fusion or fission out of the steel case, but antimatter will work. There are some practical problems with implementing this. \$\endgroup\$ – Vince Mulhollon May 10 '12 at 11:45
  • \$\begingroup\$ Lithium batteries have vents to prevent explosion when the batteries are short circuited. That's a sudden release of energy, but is not what the OP wants. \$\endgroup\$ – DanBeale May 30 '12 at 23:03

An AA battery has a finite internal resistance, which will limit the current and thus the rate at which one can remove energy. You could try transferring, over time, the energy in the battery to a lower impedance external energy store (super capacitor, inductor, flywheel, lift a rock up a cliff, etc.)

  • \$\begingroup\$ Not true. The internal resistance does not limit the current. See my answer. \$\endgroup\$ – Telaclavo May 10 '12 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.